Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: ForumUserGF on May 17, 2018, 12:39:58 AM
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I just cannot get a voltage of 1.5V as the result. Here is what I have so far:
U = E0(cathode) – E0(anode)
= 1,23 V - (-1,2 V)
= 2,43 V
But this cannot be correct...
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Well, the maths is right, but without seeing where this has come from, I'm not sure how to comment.
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Well, the maths is right, but without seeing where this has come from, I'm not sure how to comment.
Well,
Oxydation: Zn + 4OH- :rarrow: [Zn(OH)4]2- + 2e-
Reduction: 2MnO2 + H2O + 2e- :rarrow: Mn2O3 + 2OH-
And now you take E0(Mn2+/MnO2) = 1,23V and E0(Zn/Zn(OH)42-) = -1,199V. However, the result is wrong.
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If the reduction is MnO2 :rarrow: Mn2O3 why are you using E° for a different half reaction?